Find the equation of the tangent to the curve y = (5x+4)/(3x -8) at the point (2, -7)

(Much easier to explain in conversation and with drawings): But, 

An application of the quotient rule, stating that for f(x)=a(x)/b(x), f'(x) = a'(x)b(x) - a(x)b'(x) all over [b(x))]^2 Here, a(x) = 5x + 4 and b(x) = 3x - 8

Thus applying the rule we find that: f'(x) = [5(3x-8) - 3(5x + 4)] / (3x - 8)^2 As we have been given a point on this line, we can substitue in x = 2

This gives f'(x) = (-10 - 42) / 4 f'(x) = -52/4 f'(x) = -13

Given that we have the pint 2, -7 we can substitute these values, and our found gradient to work out the equation of the line at the tangent with the model of y = mx + c

-7 = -13(2) + c

c = 19

Thus the equation of the line at the tangent is y = -13x + 19

BR

Related Maths A Level answers

All answers ▸

A cubic polynomial has the form p(z)=z^3+bz^2+cz+d, z is Complex and b, c, d are Real. Given that a solution of p(z)=0 is z1=3-2i and that p(-2)=0, find the values of b, c and d.


why is sin(x) squared plus cos(x) squared 1?


How do you resolve forces on an object on an angled plane?


Find dy/dx from the equation 2xy + 3x^2 = 4y